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qtrip
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A cube with its sides numbered 1 through 6 is rolled twice 28 Apr 2012, 06:33
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C
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65% (hard)

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61% (02:11) correct 39% (02:32) wrong based on 586 sessions

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A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9

I am not sure if the OA is correct. Thanks for help !
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C
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raingary
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A cube with its sides numbered 1 through 6 is rolled twice 28 Apr 2012, 06:44
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Cube has 6 surfaces like dice . So it could be taken as a dice.

Minimum value :1 Maximum value :6
So prime no's in the range : 1+1 and 6+6 = 1 and 12
prime no : 2 ,3 , 5, 7 ,11

Different combinations for getting these values :
2 1 1
3 1 2, 2 1
5 1 4 , 4 1 ,2 3 , 3 2
7 3 4 , 4 3 , 1 6, 6 1 , 2 5, 5 2
11 5 6 , 6 5

Probability = 15/36= 5/12

Hence C is the answer.
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A cube with its sides numbered 1 through 6 is rolled twice 28 Apr 2012, 06:48
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raingary

Different combinations for getting these values :
2 1 1
3 1 2, 2 1
5 1 4 , 4 1 ,2 3 , 3 2
7 3 4 , 4 3 , 1 6, 6 1 , 2 5, 5 2
11 5 6 , 6 5

Probability = 15/36= 5/12

Hence C is the answer.
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Ok, thanks! I was counting (1,1) twice ! I agree that its 5/12 (and not 4/9)
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A cube with its sides numbered 1 through 6 is rolled twice 28 Apr 2012, 06:50
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qtrip
A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9

I am not sure if the OA is correct. Thanks for help !
Show more

Total # of outcomes is 6*6=36;

Favorable outcomes:

a-b --> prime
1-1 --> 2;

1-2 --> 3;
2-1 --> 3;

1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;

1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;

6-5 --> 11;
5-6 --> 11.

Total of 15 favorable outcomes.

P=15/36.

Answer: C.
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A cube with its sides numbered 1 through 6 is rolled twice 13 Apr 2014, 02:13
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qtrip
A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9

I am not sure if the OA is correct. Thanks for help !

Total # of outcomes is 6*6=36;

Favorable outcomes:

a-b --> prime
1-1 --> 2;

1-2 --> 3;
2-1 --> 3;

1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;

1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;

6-5 --> 11;
5-6 --> 11.

Total of 15 favorable outcomes
P=15/36.

Answer: C.
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The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?
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A cube with its sides numbered 1 through 6 is rolled twice 13 Apr 2014, 02:21
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qtrip
A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9

I am not sure if the OA is correct. Thanks for help !

Total # of outcomes is 6*6=36;

Favorable outcomes:

a-b --> prime
1-1 --> 2;

1-2 --> 3;
2-1 --> 3;

1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;

1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;

6-5 --> 11;
5-6 --> 11.

Total of 15 favorable outcomes
P=15/36.

Answer: C.


The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?
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Question says the roll landed on a and b, a and b can be different or same. The question does not give any idea about that, so assuming they are different is not correct in my opinion.
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A cube with its sides numbered 1 through 6 is rolled twice 13 Apr 2014, 06:27
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qtrip
A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability that a + b is prime?

(A) 0
(B) 1/12
(C) 5/12
(D) 7/18
(E) 4/9

I am not sure if the OA is correct. Thanks for help !

Total # of outcomes is 6*6=36;

Favorable outcomes:

a-b --> prime
1-1 --> 2;

1-2 --> 3;
2-1 --> 3;

1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;

1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;

6-5 --> 11;
5-6 --> 11.

Total of 15 favorable outcomes
P=15/36.

Answer: C.


The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?
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Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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A cube with its sides numbered 1 through 6 is rolled twice 30 Nov 2015, 19:10
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I don't know if there is a short way to solve this, but I put down everything that can be obtained from the values of a and b:
1+1 = 2 -prime
1+2=3-prime
1+3=4-not
1+4=5-prime
1+5=5-not
1+6=prime

so when a is 1, the other must be 1,2, 4, or 6. the probability that a is 1 is 1/6. the probability that b is 1,2,4,6 is 4/6 or 2/3
the first scenario when a is 1, the overall probability that the a+b is prime is 1/6 * 2/3 or 1/9

when a is 2, in order for a+b to be prime, b must be: 1,3, or 5. the probability for a to be 2 is 1/6 and the probability of b to be 1,3, or 5 is 1/2 so overall probability for a+b to be prime when a is 2 is 1/6*1/2 or 1/12

when a is 3, b must be 2 or 4 and the overall probability is 1/6 * 1/3 = 1/18

when a is 4, b must be 1 or 3 -> overall probability 1/18

when a is 5, b must be 2 or 6 -> 1/18

when a is 6, b must be 1 or 5 -> 1/18

now, the sum of all probabilities:

1/9 + 1/12 + 4(1/18)

find LCM of 9, 12, and 18 -> 36
multiply first fraction by 4, second by 3, and the last by 2.

(4+3+8)/36 or 15/36. divide by 3 both numerator and denominator -> 5/12
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A cube with its sides numbered 1 through 6 is rolled twice 28 Jun 2019, 14:30
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Rephrased, the question is asking how many ways we can get 2,3,5,7,11 from choosing 1 through 6 twice.

First, 2 ways to choose 6 numbers = 6*6 = 36 in denominator.

In numerator:
1 way, 1 choice to get a 2 {1,1}
2 ways, 1 choice to 3 {1,2} {2,1}
2 ways, 2 choices to get 5 {1,4} {4,1} & {2,3} {3,2}
2 ways, 3 choices to get 7 {3,4} {4,3} & {5,2} {2,5} & {6,1} {1,6}
2 ways, 1 choice to get a 11 {5+6} {6,5}

1+2+4+6+2 = 15

15/36 = 5/12
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A cube with its sides numbered 1 through 6 is rolled twice 02 Jul 2025, 02:10
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